JEE Mains · Physics · STD 12 -Alternating current
- JEE Mains 2025
- Easy
- MCQ
An ac current is represented as
\(\mathrm{i}=5 \sqrt{2}+10 \cos \left(650 \pi \mathrm{t}+\frac{\pi}{6}\right) \mathrm{Amp}\)
The r.m.s value of the current is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 10 Amp
Step-by-step Solution
Detailed explanation
\(i=5 \sqrt{2}+10 \cos \left(650 \pi t+\frac{\pi}{6}\right)\)
\(\mathrm{i}^2=50+100 \cos ^2\left(650 \pi \mathrm{t}+\frac{\pi}{6}\right)\)
\(+(2)(5 \sqrt{2})(10) \cos \left(650 \pi t+\frac{\pi}{6}\right)\)
\(\left\langle\mathrm{i}^2\right\rangle=50+\frac{100}{2}+0\)
\(\left. \lt \mathrm{i}^2\right\rangle=100\)
\( \lt \mathrm{i} \gt =10 \mathrm{Amp}\)
\(\mathrm{i}^2=50+100 \cos ^2\left(650 \pi \mathrm{t}+\frac{\pi}{6}\right)\)
\(+(2)(5 \sqrt{2})(10) \cos \left(650 \pi t+\frac{\pi}{6}\right)\)
\(\left\langle\mathrm{i}^2\right\rangle=50+\frac{100}{2}+0\)
\(\left. \lt \mathrm{i}^2\right\rangle=100\)
\( \lt \mathrm{i} \gt =10 \mathrm{Amp}\)
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